Q: What should I revise for H2 Physics thermodynamic systems? A: For SEAB 9478 Topic 13, revise internal energy, thermal equilibrium, work done by and on a gas, the zeroth law, the First Law (\Delta U=Q+W), specific heat capacity, and specific latent heat.
TL;DR Thermodynamics is the energy accounting chapter. Nail the distinctions between U, Q, W, then gas laws, kinetic theory, p-V graphs, and calorimetry questions become much easier to organise.
Concrete example: how to use this page
For a gas in a cylinder, ask three questions: did heat enter, did the gas do work, and did internal energy change? Answer those before substituting into the First Law.
First Law decision map
Question clue
First check
Main relation
Trap to avoid
A gas expands or is compressed
Decide whether work is done by or on the gas
ΔU=Q+W with W
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positive when work is done on the gas
Treating "work done by gas" and "work done on gas" as the same sign.
Constant-pressure expansion is stated
Convert ΔV to m3 or use kPa⋅L carefully
W=−pΔV for work done on the gas
Forgetting that 1kPa⋅L=1J.
Temperature changes without phase change
Mass, material, and ΔT are known
Q=mcΔT
Using latent heat when the state does not change.
Melting, boiling, freezing, or condensing occurs
Temperature is constant during the phase change
Q=mL
Adding a mcΔT term for the flat part of a heating curve.
Misconception check: Heat is not stored inside a system. Heating is an energy transfer across the boundary; internal energy is the store that changes after heating and work have been accounted for.
If you searched for thermal physics notes
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thermodynamics a level physics
Start with the First Law decision map, then the process constraint checkpoint.
Stay synced with the adjacent thermal-physics chapters via the H2 Physics notes hub; it bundles Temperature and Ideal Gases plus the rest of the 9478 sequence.
1 Internal energy (U)
The macroscopic state of a system fixes its internal energy: the sum of random microscopic kinetic (Ek) and potential (Ep) energies. No single thermometer can reveal U directly - you must track how energy enters or leaves.
1.1 Parent takeaway
A student who writes “heat stored in the gas” is forfeiting method marks. Heat is a transfer, not a store.
1.2 Mini-drill
State whether each contributes to U:
Scenario
Contributes to U?
Translational motion of the whole cylinder
No
Vibrations between gas molecules
Yes
2 Thermodynamic temperature (T)
Absolute temperature in kelvin is directly proportional to the mean kinetic energy per particle. Doubling T doubles ⟨Ek⟩. Always quote exam answers to 2-3 s.f. unless otherwise stated.
3 Heating (Q) vs work (W)
Symbol
Process
Positive when...
Typical formula
Q
Energy transfer by heating
Energy enters system
Q=mcΔT
W
Mechanical work
Done on system
W=−pΔV (constant p; ΔV=Vf−Vi
Sign hack: This page uses (W) as work done on the system. Work done by the gas therefore has the opposite sign in (\Delta U = Q + W).
4 Zeroth law - the thermometer principle
If A is in thermal equilibrium with B, and B with C, then A is in equilibrium with C. This justifies using a third body (a thermometer) to compare temperatures.
5 First law - the energy ledger
ΔU=Q+W.
Positive Q: heating adds energy.
Positive W: surroundings compress system.
For an ideal gas in free expansion, Q=0 and W=0, hence ΔU=0: temperature unchanged.
First Law sign checkpoint
Before substituting into ΔU=Q+W, translate the wording into this note's convention: W is positive when work is done on the system.
Wording in the question
Sign of Q
Sign of W in ΔU=Q+W
Why
Heat is supplied to the gas
Positive
Depends on expansion or compression
Energy enters by heating.
Heat is removed from the gas
Negative
Depends on expansion or compression
Energy leaves by heating.
Gas expands against external pressure
Depends on heating
Negative
The gas does work on the surroundings, so work done on the gas is negative.
Gas is compressed by the surroundings
Depends on heating
Positive
The surroundings do work on the gas.
Free expansion into a vacuum
Usually zero if insulated
Zero
No external pressure is opposed, so no mechanical work is done.
Misconception check: "work done by the gas" and "work done on the gas" have opposite signs. If you calculate work by the gas as pΔV, convert it to work on the gas by changing the sign before using this First Law form.
Process constraint checkpoint
After setting the sign convention, identify the process constraint before doing arithmetic. The constraint usually tells you which First Law term is zero or which state variable stays fixed.
Process clue
Fixed or zero quantity
First Law move
Trap to avoid
Isothermal ideal-gas process
Temperature is constant, so ΔU=0.
Set Q=−W using this note's work-on-system convention.
Saying no heat is transferred just because the temperature is constant.
Isochoric process
Volume is constant, so W=0.
Use ΔU=Q.
Calculating pΔV when ΔV=0.
Adiabatic process
Q=0.
Use ΔU=W.
Assuming temperature must stay constant; adiabatic compression can raise temperature.
Isobaric process
Pressure is constant.
Calculate work from W=−pΔV, then use ΔU=Q+W.
Treating constant pressure as zero work.
Worked check: in an insulated compression, Q=0. If the surroundings do 250J of work on the gas, then ΔU=+250J, so the gas temperature rises for an ideal gas.
Misconception check: "constant" does not always mean "zero". Constant volume gives zero work; constant temperature gives zero internal-energy change for an ideal gas; constant pressure can still involve work.
p-V area checkpoint
On a p-V diagram, the area under the path gives the work done by the gas during a volume change. Convert that area into this note's First Law sign convention before substituting into ΔU=Q+W.
Process on p-V graph
Area meaning
Sign for work done by gas
Sign of W in ΔU=Q+W
Expansion, path moves to larger V
Gas pushes surroundings through a volume increase.
Positive
Negative
Compression, path moves to smaller V
Surroundings push the gas through a volume decrease.
Negative
Positive
Constant volume
No horizontal change, so no area is swept out.
Zero
Zero
Curved or data-based path
Estimate area using rectangles or trapezia.
Depends on direction along the path.
Opposite sign to work by gas.
Worked check: a gas expands at constant pressure 120kPa from 1.0L to 3.5L. The area is 120(3.5−1.0)=300J of work done by the gas because 1kPa⋅L=1J. In ΔU=Q+W, use W=−300J.
Misconception check: the p-V area does not automatically have the sign used in every textbook convention. First decide whether it is work by the gas or work on the gas, then match the convention used in the equation.
6 Specific heat capacity (c)
Defined as the heat required per unit mass per kelvin rise:
Q=mcΔT.
6.1 Quick-check
Water has c≈4.18kJ⋅kg−1⋅K−1 - roughly 11x that of copper (≈0.39kJ⋅kg−1⋅K−1). That is why water is a coolant.
7 Specific latent heat (L)
Energy per unit mass for a phase change at constant T:
Q=mL.
Common values:\
Substance
Lf (fusion) / kJ⋅kg−1
Lv (vaporisation) / kJ⋅kg−1
Water
334
2260
8 WA & Paper 4 timing rules
Use the official paper durations and marks as a pacing guide: Papers 2 and 3 average about 1.6 min/mark, while Paper 4 averages about 3 min/mark.
Copy units first when using data-book tables.
Show full working so the energy-transfer setup, sign convention, and unit conversions remain visible.
Need structured practice on Thermodynamic Systems? Our H2 Physics tuition programme covers this topic with weekly problem sets and practical data-handling drills.
Comprehensive revision pack
9478 Section IV, Topic 13 Syllabus outcomes
Candidates should be able to:
(a) show an understanding that the macroscopic state of a system determines the internal energy of the system, and that internal energy can be expressed as the sum of a random distribution of microscopic kinetic and potential energies associated with the particles of the system.
(b) show an understanding that the thermodynamic temperature of a system is (directly) proportional to the mean microscopic kinetic energy of particles.
(c) show an understanding that when two systems are placed in thermal contact, energy is transferred (by heating) from the system at higher temperature to the system at lower temperature, until they reach the same temperature and achieve thermal equilibrium (i.e. no net energy transfer).
(d) show an understanding of the difference between the work done by a gas and the work done on a gas, and calculate the work done by a gas in expanding against a constant external pressure: W=pΔV.
(e) recall and apply the zeroth law of thermodynamics that if two systems are both in thermal equilibrium with a third system, then they are also in thermal equilibrium with each other.
(f) recall and apply the first law of thermodynamics, ΔU=Q+W, that the increase in internal energy of a system is equal to the sum of the energy transferred to the system by heating and the work done on the system.
(g) define and use the concepts of specific heat capacity and specific latent heat.
Concept map (in words)
Identify the process path (constant T, V, p or adiabatic). Track energy transfers: heating (Q) and work (W) alter internal energy (U). Use c and L for temperature changes and phase changes respectively. p-V diagrams visualise work as area under curve.
Key relations
Quantity/Process
Expression / reminder
First law
ΔU=Q+W (W positive when done on system)
Work on gas at constant pressure
W=−pΔV, where ΔV=Vf−Vi
Specific heat
Q=mcΔT
Specific latent heat
Q=mL
Isothermal ideal gas
pV=constant, ΔU=0
Isochoric process
ΔV=0⇒W=0
Adiabatic (ideal gas)
Q=0⇒ΔU=W
Internal energy of ideal gas
U=23nRT (monatomic)
Derivations & reasoning to master
First law sign convention: derive for compression and expansion, then state whether each work term is work done by the gas or work done on the gas.
Work on a p-V diagram: show W=∫pdV and interpret the area under the curve.
Heating vs temperature change: explain why adding latent heat occurs without temperature change.
Adiabatic vs isothermal: compare outcomes for ΔU to emphasise physical differences.
Worked example 1 - heating and work
An ideal gas is compressed slowly from 4.0L to 2.0L at constant pressure 150kPa while 1.6kJ of heat is removed. Find the change in internal energy.
Method: W=−pΔV (constant p); apply ΔU=Q+W. Discuss the sign of ΔU (gas cooled).
ΔV=2.0−4.0=−2.0L⇒W=−pΔV=−150(−2.0)=+300J=0.30kJ.
Heat is removed so Q=−1.6kJ. Hence
ΔU=Q+W=(−1.6+0.30)kJ=−1.3kJ.
Worked example 2 - latent heat application
0.45kg of ice at −12∘C is heated to steam at 110∘C. Calculate the total energy required. Use cice, cwater, csteam, Lf, Lv.
Approach: break into stages: warm ice → melt → warm water → vaporise → superheat steam. Sum each Q.
Using typical data-book values cice≈2.1kJ⋅kg−1⋅K−1, cwater≈4.18kJ⋅kg−1⋅K−1, csteam≈2.0kJ⋅kg−1⋅K−1, Lf=334kJ⋅kg−1, Lv=2260kJ⋅kg−1:
Parents: schedule a 1-hour Thermodynamic Systems clinic two weeks before WA 2.
Students: print the First Law equation and stick it on your water bottle - then apply it to every past-year Qn you meet.
Last updated 14 Jul 2025. Next review when SEAB publishes a newer H2 Physics syllabus document.
